Difference between revisions of "2022 AMC 12B Problems/Problem 24"

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m (Solution 1 (Complex Numbers))
 
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==Problem==
 
==Problem==
  
The figure below depicts a regular 7-gon inscribed in a unit circle.
+
The figure below depicts a regular <math>7</math>-gon inscribed in a unit circle.
 
<asy>
 
<asy>
 
         import geometry;
 
         import geometry;
Line 15: Line 15:
 
}
 
}
 
</asy>
 
</asy>
What is the sum of the 4th powers of the lengths of all 21 of its edges and diagonals?
+
What is the sum of the <math>4</math>th powers of the lengths of all <math>21</math> of its edges and diagonals?
  
 
<math>\textbf{(A) }49 \qquad \textbf{(B) }98 \qquad \textbf{(C) }147 \qquad \textbf{(D) }168 \qquad \textbf{(E) }196</math>
 
<math>\textbf{(A) }49 \qquad \textbf{(B) }98 \qquad \textbf{(C) }147 \qquad \textbf{(D) }168 \qquad \textbf{(E) }196</math>
  
==Solution (Complex numbers approach)==
+
==Solution 1 (Complex Numbers)==
  
There are 7 segments whose lengths are <math>2 \sin \frac{\pi}{7}</math>, 7 segments whose lengths are <math>2 \sin \frac{2 \pi}{7}</math>, 7 segments whose lengths are <math>2 \sin \frac{3\pi}{7}</math>.
+
There are <math>7</math> segments whose lengths are <math>2 \sin \frac{\pi}{7}</math>, <math>7</math> segments whose lengths are <math>2 \sin \frac{2 \pi}{7}</math>, <math>7</math> segments whose lengths are <math>2 \sin \frac{3\pi}{7}</math>.
  
Therefore, the sum of the 4th powers of these lengths is
+
Therefore, the sum of the <math>4</math>th powers of these lengths is
 
<cmath>
 
<cmath>
 
\begin{align*}
 
\begin{align*}
& 7 \cdot 2^4 \sin^4 \frac{\pi}{7}
+
7 \cdot 2^4 \sin^4 \frac{\pi}{7}
 
+ 7 \cdot 2^4 \sin^4 \frac{2 \pi}{7}
 
+ 7 \cdot 2^4 \sin^4 \frac{2 \pi}{7}
+ 7 \cdot 2^4 \sin^4 \frac{3 \pi}{7} \\
+
+ 7 \cdot 2^4 \sin^4 \frac{3 \pi}{7} & = \frac{7 \cdot 2^4}{(2i)^4}
& = \frac{7 \cdot 2^4}{(2i)^4}
+
\left( e^{i \frac{\pi}{7}} - e^{i \frac{-\pi}{7}} \right)^4
\left( e^{i \frac{\pi}{7}} - e^{i \frac{\pi}{7}} \right)^4
 
 
+ \frac{7 \cdot 2^4}{(2i)^4}
 
+ \frac{7 \cdot 2^4}{(2i)^4}
\left( e^{i \frac{2 \pi}{7}} - e^{i \frac{2 \pi}{7}} \right)^4
+
\left( e^{i \frac{2 \pi}{7}} - e^{i \frac{-2 \pi}{7}} \right)^4
 
+ \frac{7 \cdot 2^4}{(2i)^4}
 
+ \frac{7 \cdot 2^4}{(2i)^4}
\left( e^{i \frac{3 \pi}{7}} - e^{i \frac{4 \pi}{7}} \right)^4 \\
+
\left( e^{i \frac{3 \pi}{7}} - e^{i \frac{-3 \pi}{7}} \right)^4 \\
 
& = 7 \left(
 
& = 7 \left(
 
e^{i \frac{4 \pi}{7}} - 4 e^{i \frac{2 \pi}{7}} + 6
 
e^{i \frac{4 \pi}{7}} - 4 e^{i \frac{2 \pi}{7}} + 6
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& \quad + 7 \cdot 6 \cdot 3 \\
 
& \quad + 7 \cdot 6 \cdot 3 \\
 
& = -7 + 7 \cdot 4 + 7 \cdot 6 \cdot 3 \\
 
& = -7 + 7 \cdot 4 + 7 \cdot 6 \cdot 3 \\
& = \boxed{\textbf{(C) 147}} ,
+
& = \boxed{\textbf{(C) }147},
 
\end{align*}
 
\end{align*}
 
</cmath>
 
</cmath>
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\end{align*}
 
\end{align*}
 
</cmath>
 
</cmath>
 +
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
  
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
+
~edited by scinderella220
  
==Solution 2 (Trig approach)==
+
==Solution 2 (Trigonometry)==
  
There are 7 segments whose lengths are <math>2 \sin \frac{\pi}{7}</math>, 7 segments whose lengths are <math>2 \sin \frac{2 \pi}{7}</math>, 7 segments whose lengths are <math>2 \sin \frac{3\pi}{7}</math>.
+
There are <math>7</math> segments whose lengths are <math>2 \sin \frac{\pi}{7}</math>, <math>7</math> segments whose lengths are <math>2 \sin \frac{2 \pi}{7}</math>, <math>7</math> segments whose lengths are <math>2 \sin \frac{3\pi}{7}</math>.
  
Therefore, the sum of the 4th powers of these lengths is
+
Therefore, the sum of the <math>4</math>th powers of these lengths is
 
<cmath>
 
<cmath>
 
\begin{align*}
 
\begin{align*}
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& = 7 \cdot 2 \cdot 3 \left( 2 + 1 \right)
 
& = 7 \cdot 2 \cdot 3 \left( 2 + 1 \right)
 
- 7 \cdot 2 \left( 4 - 1 \right) \cdot \left( - \frac{1}{2} \right) \\
 
- 7 \cdot 2 \left( 4 - 1 \right) \cdot \left( - \frac{1}{2} \right) \\
& = \boxed{\textbf{(C) 147}} ,
+
& = \boxed{\textbf{(C) }147},
 
\end{align*}
 
\end{align*}
 
</cmath>
 
</cmath>
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\]
 
\]
 
</cmath>
 
</cmath>
 +
 +
For explanation see [https://artofproblemsolving.com/wiki/index.php/2022_AMC_12B_Problems/Problem_24#Supplement_.28Explanation_to_Why_cos.282.CF.80.2F7.29_.2B_cos.284.CF.80.2F7.29_.2B_cos.286.CF.80.2F7.29_.3D_-1.2F2.29 supplement].
  
 
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
 
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
  
==Solution 3==
+
==Solution 3 (Complex Numbers and Trigonometry)==
  
As explained in the first two solutions, what we are trying to find is <math>7 \cdot 2^4 \sin^4 \frac{\pi}{7} + 7 \cdot 2^4 \sin^4 \frac{2 \pi}{7} + 7 \cdot 2^4 \sin^4 \frac{3 \pi}{7}</math>. Using trig we get  
+
As explained in Solutions 1 and 2, what we are trying to find is <math>7 \cdot 2^4 \sin^4 \frac{\pi}{7} + 7 \cdot 2^4 \sin^4 \frac{2 \pi}{7} + 7 \cdot 2^4 \sin^4 \frac{3 \pi}{7}</math>. Using trig we get  
 
<cmath>
 
<cmath>
 
\begin{align*}  
 
\begin{align*}  
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\end{align*}
 
\end{align*}
 
</cmath>
 
</cmath>
Like in the second solution, we also use the fact that <math>\cos \frac{2\pi}{7} + \cos \frac{4\pi}{7} + \cos \frac{6\pi}{7} = -\frac{1}{2}</math>, which admittedly might need some explanation. Notice that  
+
Like in the second solution, we also use the fact that <math>\cos \frac{2\pi}{7} + \cos \frac{4\pi}{7} + \cos \frac{6\pi}{7} = -\frac{1}{2}</math>, which admittedly might need some explanation.  
 +
 
 +
For explanation see [https://artofproblemsolving.com/wiki/index.php/2022_AMC_12B_Problems/Problem_24#Supplement_.28Explanation_to_Why_cos.282.CF.80.2F7.29_.2B_cos.284.CF.80.2F7.29_.2B_cos.286.CF.80.2F7.29_.3D_-1.2F2.29 supplement].
 +
 
 +
Notice that  
 
<cmath>
 
<cmath>
 
\begin{align*}
 
\begin{align*}
& \cos \frac{2\pi}{7} + \cos \frac{4\pi}{7} + \cos \frac{6\pi}{7} \\
+
\cos \frac{2\pi}{7} + \cos \frac{4\pi}{7} + \cos \frac{6\pi}{7} & = \frac{1}{2}\left(e^\frac{2i\pi}{7}+ e^{-\frac{2i\pi}{7}} \right) + \frac{1}{2}\left( e^\frac{4i\pi}{7}+ e^{-\frac{4i\pi}{7}} \right) + \frac{1}{2}\left( e^\frac{6i\pi}{7}+ e^{-\frac{6i\pi}{7}}\right) \\
& = \frac{1}{2}\left(e^\frac{2i\pi}{7}+ e^{-\frac{2i\pi}{7}} \right) + \frac{1}{2}\left( e^\frac{4i\pi}{7}+ e^{-\frac{4i\pi}{7}} \right) + \frac{1}{2}\left( e^\frac{6i\pi}{7}+ e^{-\frac{6i\pi}{7}}\right) \\
 
 
& = \frac{1}{2}\left(e^\frac{2i\pi}{7}+ e^{-\frac{2i\pi}{7}}+ e^\frac{4i\pi}{7}+ e^{-\frac{4i\pi}{7}} +e^\frac{6i\pi}{7}+ e^{-\frac{6i\pi}{7}} + 1\right) - \frac{1}{2}
 
& = \frac{1}{2}\left(e^\frac{2i\pi}{7}+ e^{-\frac{2i\pi}{7}}+ e^\frac{4i\pi}{7}+ e^{-\frac{4i\pi}{7}} +e^\frac{6i\pi}{7}+ e^{-\frac{6i\pi}{7}} + 1\right) - \frac{1}{2}
 
\end{align*}
 
\end{align*}
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In the brackets we have the sum of the roots of the polynomial <math>x^7 - 1 = 0</math>. These sum to <math>0</math> by [[Vieta’s formulas]], and the desired identity follows. See [[Roots of unity]] if you have not seen this technique.
 
In the brackets we have the sum of the roots of the polynomial <math>x^7 - 1 = 0</math>. These sum to <math>0</math> by [[Vieta’s formulas]], and the desired identity follows. See [[Roots of unity]] if you have not seen this technique.
  
Going back to the question: <math>7 \cdot 2^4 \sin^4 \frac{\pi}{7} + 7 \cdot 2^4 \sin^4 \frac{2 \pi}{7} + 7 \cdot 2^4 \sin^4 \frac{3 \pi}{7} = 7 \cdot 2^4 \left(\sin^4 \frac{\pi}{7} + \sin^4 \frac{2 \pi}{7} + \sin^4 \frac{3 \pi}{7}\right) = 7 \cdot 2^4 \cdot \frac{21}{16} = \boxed{\textbf{(C) 147}}</math>.
+
Going back to the question: <cmath>7 \cdot 2^4 \sin^4 \frac{\pi}{7} + 7 \cdot 2^4 \sin^4 \frac{2 \pi}{7} + 7 \cdot 2^4 \sin^4 \frac{3 \pi}{7} = 7 \cdot 2^4 \left(\sin^4 \frac{\pi}{7} + \sin^4 \frac{2 \pi}{7} + \sin^4 \frac{3 \pi}{7}\right) = 7 \cdot 2^4 \cdot \frac{21}{16} = \boxed{\textbf{(C) }147}.</cmath>
 
~obscene_kangaroo
 
~obscene_kangaroo
  
== Solution 4 (ruler cheese) ==
+
== Solution 4 (Trigonometry) ==
 +
This solution follows the same steps as the trigonometry solutions (Solutions 2 and 3), except it gives an alternate way to prove the statement below true without complex numbers:
 +
 
 +
<cmath>\cos \frac{2\pi}{7} + \cos \frac{4\pi}{7} + \cos \frac{6\pi}{7} = -\frac{1}{2}</cmath>
 +
 
 +
<cmath>
 +
\begin{align*}
 +
S &= \cos \frac{2\pi}{7} + \cos \frac{4\pi}{7} + \cos \frac{6\pi}{7}, \\
 +
S^2 &= \cos^2 \frac{2\pi}{7} + \cos^2 \frac{4\pi}{7} + \cos^2 \frac{6\pi}{7}
 +
+ 2\cos \frac{2\pi}{7}\cos \frac{4\pi}{7} + 2\cos \frac{2\pi}{7}\cos \frac{6\pi}{7} + 2\cos \frac{4\pi}{7}\cos \frac{6\pi}{7} \\
 +
&= \left(\frac{1+ \cos \frac{4\pi}{7}}{2}\right)
 +
+ \left(\frac{1+ \cos \frac{8\pi}{7}}{2}\right)
 +
+ \left(\frac{1+ \cos \frac{12\pi}{7}}{2}\right)
 +
+ 2\cos \frac{2\pi}{7}\cos \frac{4\pi}{7} + 2\cos \frac{2\pi}{7}\cos \frac{6\pi}{7} + 2\cos \frac{4\pi}{7}\cos \frac{6\pi}{7} \\
 +
&= \frac{1}{2}(3 + S)
 +
+ \left(\cos \frac{6\pi}{7} + \cos \frac{2\pi}{7}\right)
 +
+ \left(\cos \frac{8\pi}{7} + \cos \frac{4\pi}{7}\right)
 +
+ \left(\cos \frac{10\pi}{7} + \cos \frac{2\pi}{7}\right) \\
 +
&= \frac{1}{2}(3 + S) + 2\cos \frac{2\pi}{7} + 2\cos \frac{4\pi}{7} + 2\cos \frac{6\pi}{7}\\
 +
&= \frac{1}{2}(3 + S) + 2S. \\
 +
\end{align*}
 +
</cmath>
 +
We end up with <cmath>2S^2 - 5S - 3 = 0.</cmath>
 +
Using the quadratic formula, we find the solutions for <math>S</math> to be <math>-\frac{1}{2}</math> and <math>3</math>. Because <math>3</math> is impossible, <math>S = -\frac{1}{2}</math>. 
 +
With this result, following similar to steps to Solutions 2 and 3 will get <math>\boxed{\textbf{(C) }147}</math>
 +
 
 +
~lordf
 +
 
 +
== Solution 5 (Law of Cosines) ==
 +
 
 +
Let <math>x</math>, <math>y</math>, and <math>z</math> be the lengths of the chords with arcs <math>\frac{2\pi}{7}</math>, <math>\frac{4\pi}{7}</math> and <math>\frac{6\pi}{7}</math> respectively.
 +
 
 +
Then by the law of cosines we get:
 +
<cmath>\begin{align*}
 +
x^2 &= 2\left(1-\cos\frac{2\pi}{7}\right), \\
 +
y^2 &= 2\left(1-\cos\frac{4\pi}{7}\right), \\
 +
z^2 &= 2\left(1-\cos\frac{6\pi}{7}\right).
 +
\end{align*}</cmath>
 +
The answer is then just <math>7(x^4+y^4+z^4)</math> (since there's <math>7</math> of each diagonal/side), obtained by summing the squares of the above equations and then multiplying by <math>7</math>.
 +
<cmath>\begin{align*}
 +
& \quad7\cdot2^2\left( \left(1-\cos\frac{2\pi}{7}\right)^2 + \left(1-\cos\frac{4\pi}{7}\right)^2 + \left(1-\cos\frac{6\pi}{7}\right)^2 \right) \\
 +
&= 7\cdot4\left( \left(1-2\cos\frac{2\pi}{7}+\cos^2\frac{2\pi}{7}\right) + \left(1-2\cos\frac{4\pi}{7}+\cos^2\frac{4\pi}{7}\right) + \left(1-2\cos\frac{6\pi}{7}+\cos^2\frac{6\pi}{7}\right) \right) \\
 +
&= 7\cdot4\left(3-2\left(\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}\right) + \left(\cos^2\frac{2\pi}{7}+\cos^2\frac{4\pi}{7}+\cos^2\frac{6\pi}{7}\right) \right) \\
 +
&= 7\cdot4\left(3-2\left(\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}\right) + \frac12 \left(1+\cos\frac{4\pi}{7}+1+\cos\frac{8\pi}{7}+1+\cos\frac{12\pi}{7}\right) \right) \\
 +
&= 7\cdot4\left(\frac{9}{2}-2\left(\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}\right) + \frac12 \left(\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}+\cos\frac{2\pi}{7}\right) \right) \\
 +
&= 7\cdot4\left(\frac{9}{2}-\frac{3}{2}\left(\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}\right) \right) \\
 +
&= 7\cdot4\left(\frac{9}{2}-\frac{3}{2}\left(-\frac{1}{2}\right)\right) \\
 +
&= \boxed{\textbf{(C) }147}.
 +
\end{align*}</cmath>
 +
 
 +
(Use the identity that <math> \cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7} = -\frac{1}{2}</math>.)
 +
 
 +
For explanation of this identity see [https://artofproblemsolving.com/wiki/index.php/2022_AMC_12B_Problems/Problem_24#Supplement_.28Explanation_to_Why_cos.282.CF.80.2F7.29_.2B_cos.284.CF.80.2F7.29_.2B_cos.286.CF.80.2F7.29_.3D_-1.2F2.29 supplement].
 +
 
 +
- SAHANWIJETUNGA
 +
 
 +
== Solution 6 (Ruler Measure) ==
  
 
Hope you had a ruler handy! This problem can be done with a ruler and basic estimation.
 
Hope you had a ruler handy! This problem can be done with a ruler and basic estimation.
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We know <math>\left(\frac{2.5}{2.9}\right)^4</math> is slightly less than <math>1.</math> Let's approximate it as 1 for now. Thus, <math>7\left(\frac{2.5}{2.9}\right)^4 \approx 7.</math>
 
We know <math>\left(\frac{2.5}{2.9}\right)^4</math> is slightly less than <math>1.</math> Let's approximate it as 1 for now. Thus, <math>7\left(\frac{2.5}{2.9}\right)^4 \approx 7.</math>
  
Next, <math>\left(\frac{4.5}{2.9}\right)^4</math> is slightly more than <math>\left(\frac{4.5}{3}\right)^4.</math> We know <math>\left(\frac{4.5}{3}\right)^4 = 1.5^4 = \frac{81}{16},</math> slightly more than <math>5,</math> so we can approximate <math>\left(\frac{4.5}{2.9}\right)^4</math> as <math>5.5.</math> Thus, <math>7\left(\frac{2.5}{2.9}\right)^4 \approx 38.5.</math>
+
Next, <math>\left(\frac{4.5}{2.9}\right)^4</math> is slightly more than <math>\left(\frac{4.5}{3}\right)^4.</math> We know <math>\left(\frac{4.5}{3}\right)^4 = 1.5^4 = \frac{81}{16},</math> slightly more than <math>5,</math> so we can approximate <math>\left(\frac{4.5}{2.9}\right)^4</math> as <math>5.5.</math> Thus, <math>7\left(\frac{4.5}{2.9}\right)^4 \approx 38.5.</math>
  
 
Finally, <math>\left(\frac{5.6}{2.9}\right)^4</math> is slightly less than <math>\left(\frac{5.6}{2.8}\right)^4 = 2^4 = 16.</math> We say it's around <math>15,</math> so then <math>7\left(\frac{5.6}{2.9}\right)^4 \approx 105.</math>
 
Finally, <math>\left(\frac{5.6}{2.9}\right)^4</math> is slightly less than <math>\left(\frac{5.6}{2.8}\right)^4 = 2^4 = 16.</math> We say it's around <math>15,</math> so then <math>7\left(\frac{5.6}{2.9}\right)^4 \approx 105.</math>
  
Adding what we have, we get <math>105 + 38.5 + 1 = 144.5</math> as our estimate. We see <math>\boxed{\textbf{(C)} \ 147}</math> is very close to our estimate, so we circle it and are happy that we successfully cheesed an AMC 12B problem 24.
+
Adding what we have, we get <math>105 + 38.5 + 1 = 144.5</math> as our estimate. We see <math>\boxed{\textbf{(C) }147}</math> is very close to our estimate, so we have successfully finished the problem.
  
 
~sirswagger21
 
~sirswagger21
 +
 +
== Solution 7 (Pythagorean Theorem and Trig) ==
 +
First, we put the figure in the coordinate plane with the center of the circle at the origin and a vertex on the positive x-axis. Thus, the coordinates of the vertices will be the terminal points of integer multiples of the angle <math>\frac{2\pi}{7},</math> which are <cmath>\left(\cos\dfrac{2\pi n}{7}, \sin\dfrac{2\pi n}{7}\right)</cmath> for integer <math>n.</math> Then, we notice there are three types of diagonals: the ones with chords of arcs <math>\dfrac{2\pi}{7}, \dfrac{4\pi}{7},</math> and <math>\dfrac{6\pi}{7}.</math> We notive there are here are <math>7</math> of each type of diagonal. Then, we use the [[pythagorean theorem]] to find the distance from <math>\left(\cos\frac{2\pi n}{7}, \sin\frac{2\pi n}{7}\right)</math> to <math>(1, 0)</math>:
 +
<cmath>\left(\sqrt{\left(\cos\frac{2\pi n}{7}-1\right)^2+\sin^2\frac{2\pi n}{7}}\right)^4=\left(\sqrt{\cos^2\frac{2\pi n}{7}+\sin^2\frac{2\pi n}{7}-2\cos\frac{2\pi n}{7}+1}\right)^4 </cmath>
 +
<cmath>=\left(\sqrt{2-2\cos\frac{2\pi n}{7}}\right)^4 </cmath>
 +
<cmath>=4\cos^2\frac{2\pi n}{7}-8\cos\frac{2\pi n}{7}+4.</cmath>
 +
By the cosine double angle identity, <math>\cos{2\theta}=2\cos^2\theta-1.</math> This means that <math>2\cos^2\theta=\cos{2\theta}+1.</math> Substituting this in,
 +
<cmath>4\cos^2\frac{2\pi n}{7}-8\cos\frac{2\pi n}{7}+4=2\left(\cos\frac{4\pi n}{7}+1\right)-8\cos\frac{2\pi n}{7}+4=2\cos\frac{4\pi n}{7}-8\cos\frac{2\pi n}{7}+6.</cmath> Summing this up for <math>n=1,2,3,</math>
 +
<cmath>2\cos\frac{4\pi}{7}-8\cos\frac{2\pi}{7}+6+2\cos\frac{8\pi}{7}-8\cos\frac{4\pi}{7}+6+2\cos\frac{12\pi}{7}-8\cos\frac{6\pi}{7}+6</cmath> <cmath>=2\left(\cos\frac{4\pi}{7}+\cos\frac{8\pi}{7}+\cos\frac{12\pi}{7}\right)-8\left(\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}\right)+18</cmath>
 +
<cmath>=2\left(\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}+\cos\frac{2\pi}{7}\right)-8\left(\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}\right)+18</cmath>
 +
<cmath>=2(-0.5)-8(-0.5)+18=21.</cmath>
 +
(These equalities are based on <math>\cos\theta=\cos(2\pi-\theta)</math> and <math>\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}=-\frac{1}{2}.</math>) (For explanation of this see [https://artofproblemsolving.com/wiki/index.php/2022_AMC_12B_Problems/Problem_24#Supplement_.28Explanation_to_Why_cos.282.CF.80.2F7.29_.2B_cos.284.CF.80.2F7.29_.2B_cos.286.CF.80.2F7.29_.3D_-1.2F2.29 supplement].)
 +
Finally, because there are <math>7</math> of each type of diagonal, the answer is <math>7\cdot 21=\boxed{147}.</math>
 +
 +
~[[BS2012]]
 +
 +
== Solution 8 (Roots of Unity) ==
 +
 +
Place the figure on the complex plane and let <math>w = e^{\frac{2\pi}{7}}</math> (a <math>7</math>th root of unity). The vertices of the <math>7</math>-gon are <math>w^0,w^1,w^2,\dots,w^6</math>. We wish to find
 +
<cmath>\sum_{i=0}^5\sum_{j=i+1}^6\lvert w^i-w^j\rvert^4 = \frac{1}{2}\sum_{i=0}^6\sum_{j=0}^6\lvert w^i-w^j\rvert^4.</cmath>
 +
The second expression is more convenient to work with. The factor of <math>\tfrac{1}{2}</math> is because it double-counts each edge (and includes when <math>i=j</math>, but the length is <math>0</math> so it doesn't matter).
 +
 +
Recall the identity <math>|z|^2 = z\overline{z}</math>. Since <math>|w| = 1</math>, <math>\overline{w} = w^{-1}</math>, and
 +
<cmath>\begin{align*}
 +
    \lvert w^i-w^j\rvert^4 &= ((w^i-w^j)(w^{-i}-w^{-j}))^2 \\
 +
    &= (2-w^{i-j}-w^{j-i})^2 \\
 +
    &= 4+w^{2i-2j}+w^{2j-2i}+2(-2w^{i-j}-2w^{j-i}+1) \\
 +
    &= 6-4w^{i-j}-4w^{j-i}+w^{2i-2j}+w^{2j-2i}.
 +
\end{align*}</cmath>
 +
 +
But notice that, for a fixed value of <math>i</math>, over all values of <math>j</math> from <math>0</math> to <math>6</math>, by properties of modular arithmetic with <math>w^7 = w^0</math> (or expanding),
 +
each of the <math>w^\bullet</math>-terms takes on every value in <math>w^0,w^1,w^2,\dots,w^6</math> exactly once. Since <math>\sum_{j=0}^6 w^j = 0</math>,
 +
<cmath>\begin{align*}
 +
    \frac{1}{2}\sum_{i=0}^6\sum_{j=0}^6\lvert w^i-w^j\rvert^4 &= \frac{1}{2}\sum_{i=0}^6\sum_{j=0}^6(6-4w^{i-j}-4w^{j-i}+w^{2i-2j}+w^{2j-2i}) \\
 +
    &= \frac{1}{2}\sum_{i=0}^6\left(\sum_{j=0}^6 6-4\sum_{j=0}^6 w^{i-j}-4\sum_{j=0}^6 w^{j-i}+\sum_{j=0}^6 w^{2i-2j}+\sum_{j=0}^6 w^{2j-2i}\right) \\
 +
    &= \frac{1}{2}\sum_{i=0}^6\left(\sum_{j=0}^6 6-4\sum_{j=0}^6 w^j-4\sum_{j=0}^6 w^j+\sum_{j=0}^6 w^j+\sum_{j=0}^6 w^j\right) \\
 +
    &= \frac{1}{2}\sum_{i=0}^6\sum_{j=0}^6 6 \\
 +
    &= \frac{1}{2}\cdot7\cdot7\cdot6 \\
 +
    &= \boxed{\textbf{(C) }147}.
 +
\end{align*}</cmath>
 +
 +
Fun fact: This generalizes to any regular <math>n</math>-gon with <math>n \ge 3</math> to obtain
 +
<cmath>\frac{1}{2}\sum_{i=0}^{n-1}\sum_{j=0}^{n-1}6 = 3n^2,</cmath>
 +
though for even <math>n</math>, <math>w^{2i-2j}</math> and <math>w^{2j-2i}</math> take on every value in <math>w^0,w^2,w^4,\dots,w^{n-2}</math> exactly twice. These are the <math>\tfrac{n}{2}</math>th roots of unity, so they still sum to <math>0</math>.
 +
 +
This also explains why <math>n < 3</math> doesn't follow the pattern. For <math>n = 2</math> (one edge of length <math>2</math>), since <math>w^{2i-2j} = (w^2)^{i-j} = 1^{i-j} = 1</math> and similarly for <math>w^{2j-2i}</math>, the constant term in the summation becomes <math>8</math> instead of <math>6</math>. For <math>n = 1</math> (no edges), the constant term becomes <math>0</math>.
 +
 +
-maxamc
 +
 +
~edited by [[User:emerald_block|emerald_block]]
 +
 +
== Solution 9 (Inductive Reasoning) ==
 +
 +
This is how I solve this problem:
 +
 +
It's easy to solve for <math>3</math>-gon, <math>4</math>-gon, and <math>6</math>-gon inscribed in a unit circle. (Okay, it's just the weird names for triangle, square, and hexagon)
 +
 +
For <math>3</math>-gon, the sum is equal to <math>3</math> times the <math>4</math>th power of an edge. Thus, <cmath>S_3=3\,\cdot\,\left(\sqrt{3}\right)^4=27.</cmath>
 +
 +
For <math>4</math>-gon, the sum is equal to <math>4</math> times the <math>4</math>th power of an edge, and <math>2</math> times the <math>4</math>th power of the diagonal. Thus, <cmath>S_4=4\,\cdot\,\left(\sqrt{2}\right)^4+2\,\cdot\,\left(2\right)^4=48.</cmath>
 +
 +
For <math>6</math>-gon, the sum is equal to <math>6</math> times the <math>4</math>th power of an edge, <math>6</math> times the <math>4</math>th power of the short diagonal, and <math>3</math> times the <math>4</math>th power of the long diagonal. Thus, <cmath>S_6=6\,\cdot\,\left(1\right)^4+6\,\cdot\,\left(\sqrt{3}\right)^4+3\,\cdot\,\left(2\right)^4=108.</cmath>
 +
 +
Then, I quickly noticed that <math>27=3\,\cdot\,3^2</math>, <math>48=3\,\cdot\,4^2</math>, and <math>108=3\,\cdot\,6^2</math>. So reasonably, it will work out this formula, <math>S_n=3n^2</math>. (This step is purely out of guessing, maybe have a look at Solution 8 for more info...)
 +
 +
By inductive reasoning, we got <math>S_7=3\,\cdot\,7^2=\boxed{\textbf{(C) }147}</math>.
 +
 +
- Prof. Joker
 +
 +
==Supplement (Explanation of why cos(2π/7) + cos(4π/7) + cos(6π/7) = -1/2)==
 +
 +
The sum of all roots of unity is <math>0</math>, therefore, the sum of the real parts of the roots of unity is <math>0</math>.
 +
 +
Consider the roots of unity of <math>x^n-1=0</math> where <math>n</math> is odd on the unit circle, the cosine (real parts) of all the roots under the x-axis is the same as the cosine of all the roots above the x-axis because of symmetry. Hence, <math>2</math> times the sum of the cosines of all the roots above the x-axis plus <math>1</math> is <math>0</math>.
 +
 +
In the context of this problem, <math>2(\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}) + 1 = 0</math>. Hence, <math>\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7} = -\frac{1}{2}</math>
 +
 +
~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen]
  
 
==Video Solution==
 
==Video Solution==

Latest revision as of 16:17, 2 November 2024

Problem

The figure below depicts a regular $7$-gon inscribed in a unit circle. [asy]         import geometry; unitsize(3cm); draw(circle((0,0),1),linewidth(1.5)); for (int i = 0; i < 7; ++i) {   for (int j = 0; j < i; ++j) {     draw(dir(i * 360/7) -- dir(j * 360/7),linewidth(1.5));   } } for(int i = 0; i < 7; ++i) {    dot(dir(i * 360/7),5+black); } [/asy] What is the sum of the $4$th powers of the lengths of all $21$ of its edges and diagonals?

$\textbf{(A) }49 \qquad \textbf{(B) }98 \qquad \textbf{(C) }147 \qquad \textbf{(D) }168 \qquad \textbf{(E) }196$

Solution 1 (Complex Numbers)

There are $7$ segments whose lengths are $2 \sin \frac{\pi}{7}$, $7$ segments whose lengths are $2 \sin \frac{2 \pi}{7}$, $7$ segments whose lengths are $2 \sin \frac{3\pi}{7}$.

Therefore, the sum of the $4$th powers of these lengths is \begin{align*} 7 \cdot 2^4 \sin^4 \frac{\pi}{7} + 7 \cdot 2^4 \sin^4 \frac{2 \pi}{7} + 7 \cdot 2^4 \sin^4 \frac{3 \pi}{7} & = \frac{7 \cdot 2^4}{(2i)^4} \left( e^{i \frac{\pi}{7}} - e^{i \frac{-\pi}{7}} \right)^4 + \frac{7 \cdot 2^4}{(2i)^4} \left( e^{i \frac{2 \pi}{7}} - e^{i \frac{-2 \pi}{7}} \right)^4 + \frac{7 \cdot 2^4}{(2i)^4} \left( e^{i \frac{3 \pi}{7}} - e^{i \frac{-3 \pi}{7}} \right)^4 \\ & = 7 \left( e^{i \frac{4 \pi}{7}} - 4 e^{i \frac{2 \pi}{7}} + 6 - 4 e^{- i \frac{2 \pi}{7}} + e^{- i \frac{4 \pi}{7}} \right) \\ & \quad + 7 \left( e^{i \frac{8 \pi}{7}} - 4 e^{i \frac{4 \pi}{7}} + 6 - 4 e^{- i \frac{4 \pi}{7}} + e^{- i \frac{8 \pi}{7}} \right) \\ & \quad + 7 \left( e^{i \frac{12 \pi}{7}} - 4 e^{i \frac{6 \pi}{7}} + 6 - 4 e^{- i \frac{6 \pi}{7}} + e^{- i \frac{12 \pi}{7}} \right) \\ & = 7 \left( e^{i \frac{4 \pi}{7}} + e^{i \frac{8 \pi}{7}} + e^{i \frac{12 \pi}{7}} + e^{-i \frac{4 \pi}{7}} + e^{-i \frac{8 \pi}{7}} + e^{-i \frac{12 \pi}{7}} \right) \\ & \quad - 7 \cdot 4 \left( e^{i \frac{2 \pi}{7}} + e^{i \frac{4 \pi}{7}} + e^{i \frac{6 \pi}{7}} + e^{-i \frac{2 \pi}{7}} + e^{-i \frac{4 \pi}{7}} + e^{-i \frac{6 \pi}{7}} \right) \\ & \quad + 7 \cdot 6 \cdot 3 \\ & = 7 \left( e^{i \frac{4 \pi}{7}} + e^{-i \frac{6 \pi}{7}} + e^{-i \frac{2 \pi}{7}} + e^{-i \frac{4 \pi}{7}} + e^{i \frac{6 \pi}{7}} + e^{i \frac{2 \pi}{7}} \right) \\ & \quad - 7 \cdot 4 \left( e^{i \frac{2 \pi}{7}} + e^{i \frac{4 \pi}{7}} + e^{i \frac{6 \pi}{7}} + e^{-i \frac{2 \pi}{7}} + e^{-i \frac{4 \pi}{7}} + e^{-i \frac{6 \pi}{7}} \right) \\ & \quad + 7 \cdot 6 \cdot 3 \\ & = -7 + 7 \cdot 4 + 7 \cdot 6 \cdot 3 \\ & = \boxed{\textbf{(C) }147}, \end{align*} where the fourth from the last equality follows from the property that \begin{align*} e^{i \frac{2 \pi}{7}} + e^{i \frac{4 \pi}{7}} + e^{i \frac{6 \pi}{7}} + e^{-i \frac{2 \pi}{7}} + e^{-i \frac{4 \pi}{7}} + e^{-i \frac{6 \pi}{7}} & = e^{-i \frac{6 \pi}{7}} \sum_{j=0}^6 e^{i \frac{2 \pi j}{7}} - 1  \\ & = 0 - 1 \\ & = -1 . \end{align*} ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

~edited by scinderella220

Solution 2 (Trigonometry)

There are $7$ segments whose lengths are $2 \sin \frac{\pi}{7}$, $7$ segments whose lengths are $2 \sin \frac{2 \pi}{7}$, $7$ segments whose lengths are $2 \sin \frac{3\pi}{7}$.

Therefore, the sum of the $4$th powers of these lengths is \begin{align*} & 7 \cdot 2^4 \sin^4 \frac{\pi}{7} + 7 \cdot 2^4 \sin^4 \frac{2 \pi}{7} + 7 \cdot 2^4 \sin^4 \frac{3 \pi}{7} \\ & = 7 \cdot 2^4 \left( \frac{1 - \cos \frac{2 \pi}{7}}{2} \right)^2 + 7 \cdot 2^4 \left( \frac{1 - \cos \frac{4 \pi}{7}}{2} \right)^2 + 7 \cdot 2^4 \left( \frac{1 - \cos \frac{6 \pi}{7}}{2} \right)^2 \\ & = 7 \cdot 2^2 \left( 1 - 2 \cos \frac{2 \pi}{7} + \cos^2 \frac{2 \pi}{7} \right) + 7 \cdot 2^2 \left( 1 - 2 \cos \frac{4 \pi}{7} + \cos^2 \frac{4 \pi}{7} \right) + 7 \cdot 2^2 \left( 1 - 2 \cos \frac{6 \pi}{7} + \cos^2 \frac{6 \pi}{7} \right) \\ & = 7 \cdot 2^2 \cdot 3 - 7 \cdot 2^3 \left( \cos \frac{2 \pi}{7} + \cos \frac{4 \pi}{7} + \cos \frac{6 \pi}{7} \right) + 7 \cdot 2^2 \left( \cos^2 \frac{2 \pi}{7} + \cos^2 \frac{4 \pi}{7}  + \cos^2 \frac{6 \pi}{7} \right) \\ & = 7 \cdot 2^2 \cdot 3 - 7 \cdot 2^3 \left( \cos \frac{2 \pi}{7} + \cos \frac{4 \pi}{7} + \cos \frac{6 \pi}{7} \right) + 7 \cdot 2^2 \left( \frac{1 + \cos \frac{4 \pi}{7} }{2} + \frac{1 + \cos \frac{8 \pi}{7} }{2} + \frac{1 + \cos \frac{12 \pi}{7} }{2} \right) \\ & = 7 \cdot 2^2 \cdot 3 - 7 \cdot 2^3 \left( \cos \frac{2 \pi}{7} + \cos \frac{4 \pi}{7} + \cos \frac{6 \pi}{7} \right) + 7 \cdot 2 \cdot 3 + 7 \cdot 2 \left( \cos \frac{4 \pi}{7} + \cos \frac{8 \pi}{7} + \cos \frac{12 \pi}{7} \right) \\ & = 7 \cdot 2^2 \cdot 3 - 7 \cdot 2^3 \left( \cos \frac{2 \pi}{7} + \cos \frac{4 \pi}{7} + \cos \frac{6 \pi}{7} \right) + 7 \cdot 2 \cdot 3 + 7 \cdot 2 \left( \cos \frac{4 \pi}{7} + \cos \frac{6 \pi}{7} + \cos \frac{2 \pi}{7} \right) \\ & = 7 \cdot 2 \cdot 3 \left( 2 + 1 \right) - 7 \cdot 2 \left( 4 - 1 \right) \left( \cos \frac{4 \pi}{7} + \cos \frac{6 \pi}{7} + \cos \frac{2 \pi}{7} \right) \\ & = 7 \cdot 2 \cdot 3 \left( 2 + 1 \right) - 7 \cdot 2 \left( 4 - 1 \right) \cdot \left( - \frac{1}{2} \right) \\ & = \boxed{\textbf{(C) }147}, \end{align*} where the second from the last equality follows from the property that \[ \cos \frac{4 \pi}{7} + \cos \frac{6 \pi}{7} + \cos \frac{2 \pi}{7} = - \frac{1}{2} . \]

For explanation see supplement.

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

Solution 3 (Complex Numbers and Trigonometry)

As explained in Solutions 1 and 2, what we are trying to find is $7 \cdot 2^4 \sin^4 \frac{\pi}{7} + 7 \cdot 2^4 \sin^4 \frac{2 \pi}{7} + 7 \cdot 2^4 \sin^4 \frac{3 \pi}{7}$. Using trig we get \begin{align*}  & \sin^4 \frac{\pi}{7} + \sin^4 \frac{2 \pi}{7} + \sin^4 \frac{3 \pi}{7} \\ = & \sin^2 \frac{\pi}{7} \left(1 - \cos^2 \frac{\pi}{7} \right) + \sin^2 \frac{2\pi}{7} \left(1 - \cos^2 \frac{2\pi}{7} \right) + \sin^2 \frac{3\pi}{7} \left(1 - \cos^2 \frac{3\pi}{7} \right) \\ = & \sin^2 \frac{\pi}{7} - \left(\frac{1}{2} \sin \frac{2\pi}{7}\right)^2 + \sin^2 \frac{2\pi}{7} - \left(\frac{1}{2} \sin \frac{4\pi}{7}\right)^2 + \sin^2 \frac{3\pi}{7} - \left(\frac{1}{2} \sin \frac{6\pi}{7}\right)^2\\ = & \sin^2 \frac{\pi}{7} - \frac{1}{4} \sin^2 \frac{2\pi}{7} + \sin^2 \frac{2\pi}{7} - \frac{1}{4} \sin^2 \frac{4\pi}{7} + \sin^2 \frac{3\pi}{7} - \frac{1}{4} \sin^2 \frac{6\pi}{7} \\ = & \frac{3}{4} \left(\sin^2 \frac{\pi}{7} + \sin^2 \frac{2\pi}{7} + \sin^2 \frac{3\pi}{7}\right) \\ = & \frac{3}{4} \cdot \frac{1}{2} \left(1 - \cos \frac{2\pi}{7} + 1 - \cos \frac{4\pi}{7} + 1 - \cos \frac{6\pi}{7} \right)\\ = & \frac{3}{4} \cdot \frac{1}{2} \left(3 - \left(-\frac{1}{2}\right)\right) \\ = & \frac{21}{16}. \end{align*} Like in the second solution, we also use the fact that $\cos \frac{2\pi}{7} + \cos \frac{4\pi}{7} + \cos \frac{6\pi}{7} = -\frac{1}{2}$, which admittedly might need some explanation.

For explanation see supplement.

Notice that \begin{align*} \cos \frac{2\pi}{7} + \cos \frac{4\pi}{7} + \cos \frac{6\pi}{7} & = \frac{1}{2}\left(e^\frac{2i\pi}{7}+ e^{-\frac{2i\pi}{7}} \right) + \frac{1}{2}\left( e^\frac{4i\pi}{7}+ e^{-\frac{4i\pi}{7}} \right) + \frac{1}{2}\left( e^\frac{6i\pi}{7}+ e^{-\frac{6i\pi}{7}}\right) \\ & = \frac{1}{2}\left(e^\frac{2i\pi}{7}+ e^{-\frac{2i\pi}{7}}+ e^\frac{4i\pi}{7}+ e^{-\frac{4i\pi}{7}} +e^\frac{6i\pi}{7}+ e^{-\frac{6i\pi}{7}} + 1\right) - \frac{1}{2} \end{align*} In the brackets we have the sum of the roots of the polynomial $x^7 - 1 = 0$. These sum to $0$ by Vieta’s formulas, and the desired identity follows. See Roots of unity if you have not seen this technique.

Going back to the question: \[7 \cdot 2^4 \sin^4 \frac{\pi}{7} + 7 \cdot 2^4 \sin^4 \frac{2 \pi}{7} + 7 \cdot 2^4 \sin^4 \frac{3 \pi}{7} = 7 \cdot 2^4 \left(\sin^4 \frac{\pi}{7} + \sin^4 \frac{2 \pi}{7} + \sin^4 \frac{3 \pi}{7}\right) = 7 \cdot 2^4 \cdot \frac{21}{16} = \boxed{\textbf{(C) }147}.\] ~obscene_kangaroo

Solution 4 (Trigonometry)

This solution follows the same steps as the trigonometry solutions (Solutions 2 and 3), except it gives an alternate way to prove the statement below true without complex numbers:

\[\cos \frac{2\pi}{7} + \cos \frac{4\pi}{7} + \cos \frac{6\pi}{7} = -\frac{1}{2}\]

\begin{align*} S &= \cos \frac{2\pi}{7} + \cos \frac{4\pi}{7} + \cos \frac{6\pi}{7}, \\ S^2 &= \cos^2 \frac{2\pi}{7} + \cos^2 \frac{4\pi}{7} + \cos^2 \frac{6\pi}{7} + 2\cos \frac{2\pi}{7}\cos \frac{4\pi}{7} + 2\cos \frac{2\pi}{7}\cos \frac{6\pi}{7} + 2\cos \frac{4\pi}{7}\cos \frac{6\pi}{7} \\ &= \left(\frac{1+ \cos \frac{4\pi}{7}}{2}\right) + \left(\frac{1+ \cos \frac{8\pi}{7}}{2}\right) + \left(\frac{1+ \cos \frac{12\pi}{7}}{2}\right) + 2\cos \frac{2\pi}{7}\cos \frac{4\pi}{7} + 2\cos \frac{2\pi}{7}\cos \frac{6\pi}{7} + 2\cos \frac{4\pi}{7}\cos \frac{6\pi}{7} \\ &= \frac{1}{2}(3 + S)  + \left(\cos \frac{6\pi}{7} + \cos \frac{2\pi}{7}\right) + \left(\cos \frac{8\pi}{7} + \cos \frac{4\pi}{7}\right) + \left(\cos \frac{10\pi}{7} + \cos \frac{2\pi}{7}\right) \\ &= \frac{1}{2}(3 + S) + 2\cos \frac{2\pi}{7} + 2\cos \frac{4\pi}{7} + 2\cos \frac{6\pi}{7}\\ &= \frac{1}{2}(3 + S) + 2S. \\ \end{align*} We end up with \[2S^2 - 5S - 3 = 0.\] Using the quadratic formula, we find the solutions for $S$ to be $-\frac{1}{2}$ and $3$. Because $3$ is impossible, $S = -\frac{1}{2}$. With this result, following similar to steps to Solutions 2 and 3 will get $\boxed{\textbf{(C) }147}$

~lordf

Solution 5 (Law of Cosines)

Let $x$, $y$, and $z$ be the lengths of the chords with arcs $\frac{2\pi}{7}$, $\frac{4\pi}{7}$ and $\frac{6\pi}{7}$ respectively.

Then by the law of cosines we get: \begin{align*} x^2 &= 2\left(1-\cos\frac{2\pi}{7}\right), \\ y^2 &= 2\left(1-\cos\frac{4\pi}{7}\right), \\ z^2 &= 2\left(1-\cos\frac{6\pi}{7}\right). \end{align*} The answer is then just $7(x^4+y^4+z^4)$ (since there's $7$ of each diagonal/side), obtained by summing the squares of the above equations and then multiplying by $7$. \begin{align*} & \quad7\cdot2^2\left( \left(1-\cos\frac{2\pi}{7}\right)^2 + \left(1-\cos\frac{4\pi}{7}\right)^2 + \left(1-\cos\frac{6\pi}{7}\right)^2 \right) \\ &= 7\cdot4\left( \left(1-2\cos\frac{2\pi}{7}+\cos^2\frac{2\pi}{7}\right) + \left(1-2\cos\frac{4\pi}{7}+\cos^2\frac{4\pi}{7}\right) + \left(1-2\cos\frac{6\pi}{7}+\cos^2\frac{6\pi}{7}\right) \right) \\ &= 7\cdot4\left(3-2\left(\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}\right) + \left(\cos^2\frac{2\pi}{7}+\cos^2\frac{4\pi}{7}+\cos^2\frac{6\pi}{7}\right) \right) \\ &= 7\cdot4\left(3-2\left(\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}\right) + \frac12 \left(1+\cos\frac{4\pi}{7}+1+\cos\frac{8\pi}{7}+1+\cos\frac{12\pi}{7}\right) \right) \\ &= 7\cdot4\left(\frac{9}{2}-2\left(\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}\right) + \frac12 \left(\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}+\cos\frac{2\pi}{7}\right) \right) \\ &= 7\cdot4\left(\frac{9}{2}-\frac{3}{2}\left(\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}\right) \right) \\ &= 7\cdot4\left(\frac{9}{2}-\frac{3}{2}\left(-\frac{1}{2}\right)\right) \\ &= \boxed{\textbf{(C) }147}. \end{align*}

(Use the identity that $\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7} = -\frac{1}{2}$.)

For explanation of this identity see supplement.

- SAHANWIJETUNGA

Solution 6 (Ruler Measure)

Hope you had a ruler handy! This problem can be done with a ruler and basic estimation.

First, measuring the radius of the circle obtains $2.9$ cm (when done on the paper version). Thus, any other measurement we get for the sides/diagonals should be divided by $2.9$.

Measuring the sides of the circle gets $2.5$ cm. The shorter diagonals are $4.5$ cm, and the longest diagonals measure $5.6$ cm. Thus, we'd like to estimate \[7\left(\frac{2.5}{2.9}\right)^4 + 7\left(\frac{4.5}{2.9}\right)^4 + 7\left(\frac{5.6}{2.9}\right)^4.\]

We know $\left(\frac{2.5}{2.9}\right)^4$ is slightly less than $1.$ Let's approximate it as 1 for now. Thus, $7\left(\frac{2.5}{2.9}\right)^4 \approx 7.$

Next, $\left(\frac{4.5}{2.9}\right)^4$ is slightly more than $\left(\frac{4.5}{3}\right)^4.$ We know $\left(\frac{4.5}{3}\right)^4 = 1.5^4 = \frac{81}{16},$ slightly more than $5,$ so we can approximate $\left(\frac{4.5}{2.9}\right)^4$ as $5.5.$ Thus, $7\left(\frac{4.5}{2.9}\right)^4 \approx 38.5.$

Finally, $\left(\frac{5.6}{2.9}\right)^4$ is slightly less than $\left(\frac{5.6}{2.8}\right)^4 = 2^4 = 16.$ We say it's around $15,$ so then $7\left(\frac{5.6}{2.9}\right)^4 \approx 105.$

Adding what we have, we get $105 + 38.5 + 1 = 144.5$ as our estimate. We see $\boxed{\textbf{(C) }147}$ is very close to our estimate, so we have successfully finished the problem.

~sirswagger21

Solution 7 (Pythagorean Theorem and Trig)

First, we put the figure in the coordinate plane with the center of the circle at the origin and a vertex on the positive x-axis. Thus, the coordinates of the vertices will be the terminal points of integer multiples of the angle $\frac{2\pi}{7},$ which are \[\left(\cos\dfrac{2\pi n}{7}, \sin\dfrac{2\pi n}{7}\right)\] for integer $n.$ Then, we notice there are three types of diagonals: the ones with chords of arcs $\dfrac{2\pi}{7}, \dfrac{4\pi}{7},$ and $\dfrac{6\pi}{7}.$ We notive there are here are $7$ of each type of diagonal. Then, we use the pythagorean theorem to find the distance from $\left(\cos\frac{2\pi n}{7}, \sin\frac{2\pi n}{7}\right)$ to $(1, 0)$: \[\left(\sqrt{\left(\cos\frac{2\pi n}{7}-1\right)^2+\sin^2\frac{2\pi n}{7}}\right)^4=\left(\sqrt{\cos^2\frac{2\pi n}{7}+\sin^2\frac{2\pi n}{7}-2\cos\frac{2\pi n}{7}+1}\right)^4\] \[=\left(\sqrt{2-2\cos\frac{2\pi n}{7}}\right)^4\] \[=4\cos^2\frac{2\pi n}{7}-8\cos\frac{2\pi n}{7}+4.\] By the cosine double angle identity, $\cos{2\theta}=2\cos^2\theta-1.$ This means that $2\cos^2\theta=\cos{2\theta}+1.$ Substituting this in, \[4\cos^2\frac{2\pi n}{7}-8\cos\frac{2\pi n}{7}+4=2\left(\cos\frac{4\pi n}{7}+1\right)-8\cos\frac{2\pi n}{7}+4=2\cos\frac{4\pi n}{7}-8\cos\frac{2\pi n}{7}+6.\] Summing this up for $n=1,2,3,$ \[2\cos\frac{4\pi}{7}-8\cos\frac{2\pi}{7}+6+2\cos\frac{8\pi}{7}-8\cos\frac{4\pi}{7}+6+2\cos\frac{12\pi}{7}-8\cos\frac{6\pi}{7}+6\] \[=2\left(\cos\frac{4\pi}{7}+\cos\frac{8\pi}{7}+\cos\frac{12\pi}{7}\right)-8\left(\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}\right)+18\] \[=2\left(\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}+\cos\frac{2\pi}{7}\right)-8\left(\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}\right)+18\] \[=2(-0.5)-8(-0.5)+18=21.\] (These equalities are based on $\cos\theta=\cos(2\pi-\theta)$ and $\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}=-\frac{1}{2}.$) (For explanation of this see supplement.) Finally, because there are $7$ of each type of diagonal, the answer is $7\cdot 21=\boxed{147}.$

~BS2012

Solution 8 (Roots of Unity)

Place the figure on the complex plane and let $w = e^{\frac{2\pi}{7}}$ (a $7$th root of unity). The vertices of the $7$-gon are $w^0,w^1,w^2,\dots,w^6$. We wish to find \[\sum_{i=0}^5\sum_{j=i+1}^6\lvert w^i-w^j\rvert^4 = \frac{1}{2}\sum_{i=0}^6\sum_{j=0}^6\lvert w^i-w^j\rvert^4.\] The second expression is more convenient to work with. The factor of $\tfrac{1}{2}$ is because it double-counts each edge (and includes when $i=j$, but the length is $0$ so it doesn't matter).

Recall the identity $|z|^2 = z\overline{z}$. Since $|w| = 1$, $\overline{w} = w^{-1}$, and \begin{align*}     \lvert w^i-w^j\rvert^4 &= ((w^i-w^j)(w^{-i}-w^{-j}))^2 \\     &= (2-w^{i-j}-w^{j-i})^2 \\     &= 4+w^{2i-2j}+w^{2j-2i}+2(-2w^{i-j}-2w^{j-i}+1) \\     &= 6-4w^{i-j}-4w^{j-i}+w^{2i-2j}+w^{2j-2i}. \end{align*}

But notice that, for a fixed value of $i$, over all values of $j$ from $0$ to $6$, by properties of modular arithmetic with $w^7 = w^0$ (or expanding), each of the $w^\bullet$-terms takes on every value in $w^0,w^1,w^2,\dots,w^6$ exactly once. Since $\sum_{j=0}^6 w^j = 0$, \begin{align*}     \frac{1}{2}\sum_{i=0}^6\sum_{j=0}^6\lvert w^i-w^j\rvert^4 &= \frac{1}{2}\sum_{i=0}^6\sum_{j=0}^6(6-4w^{i-j}-4w^{j-i}+w^{2i-2j}+w^{2j-2i}) \\     &= \frac{1}{2}\sum_{i=0}^6\left(\sum_{j=0}^6 6-4\sum_{j=0}^6 w^{i-j}-4\sum_{j=0}^6 w^{j-i}+\sum_{j=0}^6 w^{2i-2j}+\sum_{j=0}^6 w^{2j-2i}\right) \\     &= \frac{1}{2}\sum_{i=0}^6\left(\sum_{j=0}^6 6-4\sum_{j=0}^6 w^j-4\sum_{j=0}^6 w^j+\sum_{j=0}^6 w^j+\sum_{j=0}^6 w^j\right) \\     &= \frac{1}{2}\sum_{i=0}^6\sum_{j=0}^6 6 \\     &= \frac{1}{2}\cdot7\cdot7\cdot6 \\     &= \boxed{\textbf{(C) }147}. \end{align*}

Fun fact: This generalizes to any regular $n$-gon with $n \ge 3$ to obtain \[\frac{1}{2}\sum_{i=0}^{n-1}\sum_{j=0}^{n-1}6 = 3n^2,\] though for even $n$, $w^{2i-2j}$ and $w^{2j-2i}$ take on every value in $w^0,w^2,w^4,\dots,w^{n-2}$ exactly twice. These are the $\tfrac{n}{2}$th roots of unity, so they still sum to $0$.

This also explains why $n < 3$ doesn't follow the pattern. For $n = 2$ (one edge of length $2$), since $w^{2i-2j} = (w^2)^{i-j} = 1^{i-j} = 1$ and similarly for $w^{2j-2i}$, the constant term in the summation becomes $8$ instead of $6$. For $n = 1$ (no edges), the constant term becomes $0$.

-maxamc

~edited by emerald_block

Solution 9 (Inductive Reasoning)

This is how I solve this problem:

It's easy to solve for $3$-gon, $4$-gon, and $6$-gon inscribed in a unit circle. (Okay, it's just the weird names for triangle, square, and hexagon)

For $3$-gon, the sum is equal to $3$ times the $4$th power of an edge. Thus, \[S_3=3\,\cdot\,\left(\sqrt{3}\right)^4=27.\]

For $4$-gon, the sum is equal to $4$ times the $4$th power of an edge, and $2$ times the $4$th power of the diagonal. Thus, \[S_4=4\,\cdot\,\left(\sqrt{2}\right)^4+2\,\cdot\,\left(2\right)^4=48.\]

For $6$-gon, the sum is equal to $6$ times the $4$th power of an edge, $6$ times the $4$th power of the short diagonal, and $3$ times the $4$th power of the long diagonal. Thus, \[S_6=6\,\cdot\,\left(1\right)^4+6\,\cdot\,\left(\sqrt{3}\right)^4+3\,\cdot\,\left(2\right)^4=108.\]

Then, I quickly noticed that $27=3\,\cdot\,3^2$, $48=3\,\cdot\,4^2$, and $108=3\,\cdot\,6^2$. So reasonably, it will work out this formula, $S_n=3n^2$. (This step is purely out of guessing, maybe have a look at Solution 8 for more info...)

By inductive reasoning, we got $S_7=3\,\cdot\,7^2=\boxed{\textbf{(C) }147}$.

- Prof. Joker

Supplement (Explanation of why cos(2π/7) + cos(4π/7) + cos(6π/7) = -1/2)

The sum of all roots of unity is $0$, therefore, the sum of the real parts of the roots of unity is $0$.

Consider the roots of unity of $x^n-1=0$ where $n$ is odd on the unit circle, the cosine (real parts) of all the roots under the x-axis is the same as the cosine of all the roots above the x-axis because of symmetry. Hence, $2$ times the sum of the cosines of all the roots above the x-axis plus $1$ is $0$.

In the context of this problem, $2(\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}) + 1 = 0$. Hence, $\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7} = -\frac{1}{2}$

~isabelchen

Video Solution

https://youtu.be/nO5p_xfXykI

~ ThePuzzlr

https://youtu.be/yRbweIYtLU8

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

See Also

2022 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 23
Followed by
Problem 25
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All AMC 12 Problems and Solutions

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